answers to chapter 4 in-chapter problems
TRANSCRIPT
Answers To Chapter 4 In-Chapter Problems.
4.1. The numbering of the atoms is quite difficult in this problem. The number of Me groups in the
product suggests that at least two equivalents of the bromide are incorporated into the product. But which
ring atoms are C3 and which one is C6? And even if one of the ring carbons is arbitrarily chosen as C6,
there is still the question of whether C3 or C2 becomes attached to C6. This problem is solved by noting
that step 1 turns the bromide into a Grignard reagent, which is nucleophilic at C3, so it is likely to attack
C6, and electrophilic atom. Make: C3–C6, C3´–C6, C2–C2´. Break: C6–O7, C6–O8, C3–Br, C3´–Br.
MeBr
Me
Me H
Me
Me Me
Me1 23
4
Mg CH3COEt
O
H+5
6
78 4
3
21
4´3´
2´1´
5
6
In the first step, the bromide is converted to a Grignard reagent. In the second step, two equivalents of the
Grignard reagent react with the ester by addition–elimination–addition. (Remember, the ketone that is
initially obtained from reaction of a Grignard reagent with an ester by addition–elimination is more
electrophilic than the starting ester, so addition of a second Grignard reagent to the ketone to give an
alcohol is faster than the original addition to give the ketone.) In the last step, addition of acid to the
tertiary, doubly allylic alcohol gives a pentadienyl cation that undergoes electrocyclic ring closure. Loss of
H+ gives the observed product.
MeBr
Me
MgMe
MgBr
Me
CH3COEt
O
Me
Me
OEt
O MeH H H
Me
Me
O Me
Me
Me
Me
O MeBrMg
Me
Me
MeH+
H
H
H H
Me
H2O Me
Me
Me Me
Me Me
Me Me
Me
H H HH
Me Me
Me Me
Me
HH
product
4.2. As usual, the key to this problem is numbering correctly. The main question is whether the ester C
in the product is C3 or C4. Because a ring contraction from 6- to 5-membered is likely to proceed by a
Favorskii rearrangement, where the last step is cleavage of a cyclopropanone, it makes sense to label the
Chapter 4 2
ester C as C4. Make: C3–C5, C4–C15. Break: C3–O9, C4–C5, C5–O13, O11–Ts12.
OO
TsO OTs
OTs
NaOMeO
O
12
34
5
6
78
OMe
O910
11 12
13 14
1512
78
6
5
15
OO
OMe
O
12
78
6
5
15
OO
OMeO
2
3
45
6
4
11
3HENCE
78
15
Hold on! If the O11–Ts12 bond is broken, and the electrons go to O (as seems reasonable), what happens
to the Ts? Some nucleophile must form a bond to it. The only nucleophile in the mixture is MeO–, so let’s
add Ts12–O15 to our make list.
NaOMe is a good base, and with all these TsO groups, an E2 elimination reaction to break a C–OTs bond
seems reasonable. Either the C3–O9 or the C5–O13 bond can be cleaved; we choose the C5–O13 bond
here, but cleavage of the other bond works, too. The product is an enol tosylate. A second elimination
reaction is not possible, but at this point we can form the Ts12–O15 bond and cleave the O11–Ts12 bond
by having MeO– attack Ts 12, displacing O11 to make an enolate. Electrocyclic ring closing with
concurrent cleavage of the C3–O9 bond gives a cyclopropanone. Addition of O15 to C4 and then C4–C5
bond cleavage with concurrent protonation of C5 by solvent gives the product.
OO
TsO OTs
OTs
HH
H
–OMeO
O
TsO
O
H
H
Ts
–OMeO
O
TsO
O
H
H
OO
O
HH
–OMeO
O
O
HHMeO
H OMeO
O
O
HHMeO
H
4.3. Make: C1–C8. Break: none.
O
H3C CH3CH3H3C
NaOH
O
H3C CH3
CH3123
45
6
7
8
9
10
1
23
45
6
78
9
10
HH HH
H
HH
Deprotonation of C1 gives an enolate ion, which in this compound is actually a 1,3,5-hexatriene. As such
it can undergo an electrocyclic ring closing. Protonation gives the product.
Chapter 4 3
O
H3CH2CCH3H3C
H
OH
O
H3C CH2CH3H3C
O
H3C CH2
CH3CH3
productOH
HHH H H H H
You may have been tempted to draw the C1–C8 bond-forming reaction as a conjugate addition. However,
once C1 is deprotonated, the carbonyl group is no longer electrophilic, because it is busy stabilizing the
enolate. It is much more proper to think of the bond-forming reaction as an electrocyclic ring closure.
This problem illustrates why it is so important to consider all the resonance structures of any species.
4.4. Make: C2–C7. Break: C2–C5.
OOH
∆H
H HH
H
H
H
H
H
H
H
1
23
4 56 7
1
23
45
6
7
You may be very tempted to draw the following mechanism for the reaction:
OH
H HH
H
H
OH
H HH
H
H~H+
OH
HH H
H
H
However, this mechanism is not correct. It is a [1,3]-sigmatropic rearrangement, and for reasons which
are discussed in Section 4.4.2, [1,3]-sigmatropic rearrangements are very rare under thermal conditions.
A much better mechanism can be written. The C2–C5 bond is part of a cyclobutene, and cyclobutenes
open very readily under thermal conditions. After the electrocyclic ring opening, a 1,3,5-hexatriene is
obtained, and these compounds readily undergo electrocyclic ring closure under thermal conditions.
Tautomerization then affords the product.
OH
H HH
H
HC
OH
HH H
H
HOH
HH H
H
H~H+
product
4.5. The product has a cis ring fusion.
Chapter 4 4
H
H
HH
H
H
HH
H
H
HH
H
H
HH
6 e–
4.6. The first electrocyclic ring closure involves eight electrons, so it is conrotatory under thermal
conditions, and the two hydrogen atoms at the terminus of the tetraene, which are both in, become trans.
The second electrocyclic ring closure involves six electrons, so it is disrotatory under thermal conditions,
and the two hydrogen atoms at the terminus of the triene, which are both out, become cis. This is the
arrangement observed in the natural product.
CO2R
Ph
CO2R
Ph
H
HH
HH
H
HH
CO2R
Ph
H
HH
H H
HH
Ph
H H
H
CO2R
∆
H
4.7. The HOMO of the pentadienyl cation is ψ1, which is antisymmetric, so a conrotatory ring closure
occurs, consistent with the four electrons involved in this reaction. The HOMO of the pentadienyl anion is
ψ2, which is symmetric, so a disrotatory ring closure occurs, consistent with the six electrons involved in
this reaction.
+ + + +
+ · – –
· – · +
– · + –
+
+
+
+
– + – ++
ψ0
ψ1
ψ2
ψ3
ψ4
MOs of the pentadienyl π system
pentadienyl cation
pentadienyl anion
Chapter 4 5
pentadienyl cation HOMO H H
H H
conrotatory
H
H
H
H
pentadienyl anion HOMO H H
H H
disrotatory
H
H
H
H
4.8. Make: O1–C9, N2–C13, C10–C13. Break: C13–O14.
H
H
NHOH
CH3
CH2=O ON
CH3
12
34
5
6
7
8
910
11
12
13 14
1
2 34
5
67
89
10
111213
The new five-membered, heterocyclic ring clues you in to the fact that a 1,3-dipolar cycloaddition has
occurred here to form bonds O1–C9 and C10–C13. Disconnect these bonds, putting a + charge on C13
and a – charge on O1, to see the immediate precursor to the product.
ON
CH3
1
2
910
13 ON
H2C
CH3 H
H
N
CH3
H2C O
When this disconnection is written in the forward direction along with some curved arrows, it is the last
step in the reaction. Now all you have to do is make N2–C13 and break C13–O14. This is easy to do: N2
attacks C13, proton transfer occurs, N2 expels O14, and deprotonation gives the nitrone.
H
H
NHOH
CH3
CH2=OH
H
NH
CH3
H2C
OH
O
~H+H
H
N
CH3
H2C
OH
HO
Chapter 4 6
H
H
N
CH3
H2COH
OH
~H+O
N
H2C
CH3
product
4.9. Me2S attacks one of the O atoms involved in the O–O bond, displacing O–. Hemiacetal collapse to
the carbonyl compounds then occurs.
O
O
O
R
R´SMe2 O
O
O
R
R´
SMe2
O
O
O
R
R´
SMe2
4.10. Make: C7–C2´, C9–O1´. Break: C2´–O1´.
HN
N
O
Me
O
R
HN
N
O
O
R N N
MeOH
Me
O
R
H
hνHN
N
O
Me
O
R
R= sugar phosphate backbone of DNA.
1
23
45 6H H
H7
89
1´
2´3´
4´5´ 6´ 7´
8´9´
1
23
45 6 78
9
6´4´
5´
3´
2´
1´
9´
8´ 7´
Making the C7–C2´ and C9–O1´ suggests a [2 + 2] photocycloaddition. Then the lone pair on N3´ expels
O1´ from C2´ to give the observed product (after proton transfer).
HN
N
O
Me
O
RH HN
N
O
Me
O
R
H
hν HN
N
O
Me
O
RHHN
N
O
Me
O
R
H
Chapter 4 7
HN
N
O
Me
O
RHHN
N
O
Me
O
R
H
~H+
product
4.11. The numbering is not straightforward in this reaction, but if you draw in the H atoms you can see
that the two CH groups in the new benzene ring in the product probably come from two CH groups in
norbornadiene. Atoms unaccounted for in the written product include C9 and O10 (can be lost as CO),
C17 to C21 (can be lost as cyclopentadiene), and O1 and O4 (can be lost as H2O). Make: C2–C8, C3–
C11, C8–C16, C11–C15. Break: O1–C2, C3–O4, C8–C9, C9–C11, C15–C19, C16–C17.
O
O
H3C CH3
O O O
, cat. glycine
1
2
34
5 6
7
8 9
10
11
12
13
14
15
1617
18
1920
21
2
3
56
H
HH
HH
H
H H
H H H HH
H
O CH3
O CH3
7
8
141213
11 15
16
C
O
9
10
HH
H
H H
17
18
1920
21
Glycine acts as an acid–base catalyst in this reaction. C8 and C11 are very acidic, and once deprotonated
they are very nucleophilic, so they can attack C2 and C3 in an aldol reaction. Dehydration gives a key
cyclopentadienone intermediate. (The mechanism of these steps is not written out below.)
Cyclopentadienones are antiaromatic, so they are very prone to undergo Diels–Alder reactions. Such a
reaction occurs here with norbornadiene. A retro-Diels–Alder reaction followed by a [4 + 1]
retrocycloaddition affords the product.
O
O
H3C CH3
O O O
H H H H
OH
OH
O
H3CO
H3CO
H
H
Chapter 4 8
O
H3CO
H3CO
H
H
HH
H
H H
O
H3CO
H3CO
H
H
H
H
H
H
H H
O
H3CO
H3CO
H
H
product
4.12. Make: C3–C9, C3–C14, C6–C10, C6–C13. Break: C3–N4, N5–C6.
H
H
CN
N
N
CN
CN
110 °CCN
+
123
4
H
H
H
H
H
H
HH
H
H5
67
8
9
1011 12
13
141516 1
23
7 86
9
1011
12 13
1415
16
The C3–C9 and C6–C10 bonds can be made by a Diels–Alder reaction. Then loss of N2 and cleavage of
the C3–N4 and N5–C6 bonds can occur by a retro-Diels–Alder reaction. This step regenerates a diene,
which can undergo another, intramolecular Diels–Alder reaction with the C13–C14 π bond to give the
product.
N
N
CN
CN
H
H
H
H
H
H
H
H
NN
H
H
NC
NC
H
H
H
H
H
H
NC
NC
product
H
H
4.13. The [6+4] cycloaddition involves five pairs of electrons (an odd number), so it is thermally
Chapter 4 9
allowed. The [4+3] cationic cycloaddition involves three pairs of electrons, so it is also thermally allowed.
4.14. Make: C6–C8. Break: C6–C7.
i-Pr
i-Pr
i-Pr
280 °C, 24 hi-Pr
H
H
H
H
H
H
HH
H
HH
HH
HH
H
H
H
H
H1
2
3
4
56
7
8 910 11
12
1314
1515
1413
12
111098
7
5
2
1
4
3
6
Making and breaking bonds to C6 suggests a [1,n] sigmatropic rearrangement, and a [1,5] sigmatropic
rearrangement, one of the most common types, is possible here. Once the rearrangement is drawn,
however, the mechanism is not complete, even though all bonds on the make & break list have been
crossed off. C8 still has one extra H and C9 has one too few. Both these problems can be taken care of
by another [1,5] sigmatropic rearrangement. This step, by the way, reestablishes the aromatic ring.
i-Pr
i-PrH
H
H
H
H
H
HH
H
Hi-Pr
i-PrH
H
H
H
H
H
HH
H
H
6 7
8 9
product
8
4.15. Make: C1–C9. Break: C3–N8.
N CO2EtBn
DBUN
Bn CO2Et
12
3
4
5
6
7
89
10
1
234
5
6
7
8
9
10
HH
H
H
H H
H
H
H
H
H
Deprotonation of C9 by DBU gives an ylide (has positive and negative charges on adjacent atoms that
cannot quench each other with a π bond), a compound which is particularly prone to undergo [2,3]
sigmatropic rearrangements when an allyl group is attached to the cationic center, as is the case here.
Esters are not normally acidic enough to be deprotonated by DBU, but in this ester the N+ stabilizes the
enolate by an inductive effect.
Chapter 4 10
N CO2EtBn
DBU
HH
H
H
H H
N CO2EtBn
HH
H
H
H
N CO2EtBn
HH
H
H
H
4.16. Make: C4–C10. Break: Cl1–N2.
NRMeS CO2Et
NHR
SMe
CO2EtCl
H HH
H
H
H
H
H
H
H H
H
1
23
45
6
7 89
1011
11
10
92
3
4
56
7
8
The most unusual bond in this system is the N–Cl bond. The nucleophilic substitution step must involve
cleavage of this bond. No base is present, but S is an excellent nucleophile, even in its neutral form, so
the first step probably entails formation of an S9–N2 bond. Now we have to make the C4–C10 bond and
make the S9–N2 bond. Deprotonation of C4 gives an ylide, which as discussed in problem 4.15 is likely
to undergo a [2,3] sigmatropic rearrangement. Tautomerization to rearomatize then gives the product.
NRMeS CO2Et
Cl
H HH
H
H
H
H
NR
SMe
H
H
H
H
H H
H
CO2Et
base
NR
SMe
H
H
H
H
H H
CO2Et NRMeS
H
H
H
H
H H
CO2Et~H+
product
4.17. The reaction in question is:
O∆
OH
To name the reaction, draw a dashed line where the new bond is made, draw a squiggly line across the
bond that is broken, and count the number of atoms from the termini of the dashed bond to the termini of
the squiggly bond.
Chapter 4 11
O
12
34
5
1
23
This reaction would be a [3,5] sigmatropic rearrangement, an eight-electron reaction, and hence would
require that one component be antarafacial. Not likely! A more reasonable mechanism begins with the
same [3,3] sigmatropic rearrangement that gives 2-allylphenol. However, instead of tautomerization to
give the aromatic product, a second [3,3] sigmatropic rearrangement occurs. Then tautomerization gives
the product.
O
H
H
H
H
H
H
H H
H
H
O
H
H
H
H
H H
H
H
H
H
O
H
H
H
H
H
H
H
H
H
H
~H+
product
4.18. Both the Stevens rearrangement and the nonallylic Wittig rearrangement begin with deprotonation of
the C atom next to the heteroatom followed by an anionic [1,2] sigmatropic rearrangement. Both involve
four electrons, an even number of electron pairs, and hence if either is concerted then one of the two
components of the reaction must be antarafacial. This condition is extremely difficult to fulfill, and hence
it is much more likely that both reactions are nonconcerted. Both the Stevens rearrangement and the
nonallylic Wittig rearrangement are thought to proceed by homolysis of a C–S or C–O bond and
recombination of the C radical with the neighboring C atom.
Ph OCH3
BuLiH H
Ph OCH3
H
Ph OCH3
H
Ph O–
CH3
Ph OCH3
H
4.19. Make: N1–C11, C2–C8. Break: C2–C6, C11–O12.
OH
NHMe
Ph
MeCHO
cat. H+NMe
OPh
H
Me1
23
4
5
6
7 89 10 11 12
12
3
4
5 6
7
8
910
11H2O
12
Chapter 4 12
The N1–C11 bond is easily made first. Cleavage of the C11–O12 bond gives an iminium ion that is also a
1,5-(hetero)diene. The Cope rearrangement occurs to give a new iminium ion and an enol. Attack of the
enol on the iminium ion (the Mannich reaction) affords the product.
OH
NHMe
Ph
Me OHOH
NH
Ph
Me
OH
~H+
OH
N
Ph
Me
OH2
Me Me
OH
N
Ph
Me
OH
N
Ph
Me
OH
N
Ph
Me
–H+product
Me Me Me
Now the stereochemistry. Assume the thermodynamically more stable iminium ion forms (Me groups
cis). The Cope rearrangement occurs from a chair conformation. This puts the Ph, H2, and H11 all
pointing up both before and after the rearrangement. Assuming the Mannich reaction occurs without a
change in conformation (a reasonable assumption, considering the proximity of the nucleophilic and
electrophilic centers), the Ph, H2, and H11 should all be cis in the product.
N
MeOH
Me
PhH H
N
MeOH
Me
PhH H
N
MeO
Me
PhH H
211 211
4.20. Deprotonation of one of the Me groups adjacent to S gives an ylide which can undergo a retro-
hetero-ene reaction to give the observed products.
H2C
S
CH3
OR
HHNEt3H
CH2
S
CH3
OR
HHCH3
S
CH3
OR
H
If (CD3)2SO (deuterated DMSO) is used for the Swern reaction, the E2 mechanism predicts that the sulfide
product should be (CD3)2S; the retro-hetero-ene mechanism predicts that it should be (CD3)S(CHD2).
Guess which product is actually found?
Chapter 4 13
Answers To Chapter 4 End-of-Chapter Problems.
1.
(a) An eight-electron [4+4] cycloaddition. It proceeds photochemically.
(b) A four-electron conrotatory electrocyclic ring opening. It proceeds thermally.
(c) A six-electron ene reaction. (Note the transposition of the double bond.) It proceeds thermally.
(d) A six-electron [1,5] sigmatropic rearrangement. It proceeds thermally.
(e) A ten-electron [8+2] cycloaddition. It proceeds thermally.
(f) A six-electron [2,3] sigmatropic rearrangement. It proceeds thermally.
(g) A six-electron disrotatory electrocyclic ring opening. It proceeds thermally.
(h) A four-electron disrotatory electrocyclic ring closing. It proceeds photochemically.
(i) A six-electron disrotatory electrocyclic ring closing. It proceeds thermally.
(j) A six-electron [3+2] (dipolar) cycloaddition. It proceeds thermally.
(k) A four-electron [2+2] cycloaddition. It proceeds photochemically.
(l) A six-electron conrotatory electrocyclic ring opening. It proceeds photochemically.
2.
(a) Regio: RNH and CHO are 1,2. Stereo: CHO
and CH3 remain trans; NHR is out, CHO is endo,
so they are cis in product.
BnO2CHN CHO
CH3
(b) The two CH3 groups are both out groups, so
they are cis in product.
OO
H3C CH3
(c) Regio: C4 of diene is nucleophilic, so it makes a
bond to electrophilic C of dienophile. Stereo: EtO is
out, CO2Et group is endo, so they are cis in
product.
OEtO CO2Et
(d) Regio: CHO and OSiMe3 are 1,4. Stereo: the
CH2CH2 bridge is in at both ends of the diene,
CHO is endo, so they are trans in product.
Me3SiO
H3C
CHO
CH3
Chapter 4 14
(e) Dienophile adds to less hindered face of diene.
C(sp3) of five-membered ring is in, NO2 is endo,
so they are trans in product.
NO2
HEtO
EtO
(f) Regio: Nucleophilic O adds to electrophilic β C
of unsaturated ester. Stereo: alkyl and CO2Me
groups remain trans; H is in, CO2Me is endo, so
they are trans in product.
N O
CO2MeH
OMs
(g) Stereo: CO2Me groups remain trans. Ar group
is probably out for steric reasons, CO2Me is endo,
so the two are cis in the product.
N
CO2MeMeO2C
ArPh
(h) The [14+2] cycloaddition must be antarafacial
with respect to one component. The two in groups
of the 14-atom component become trans in the
product.
CN
CNNC
NC
H H
3. 1,3,5,7-Cyclononatetraene can theoretically undergo three different electrocyclic ring closures.
HH
8 e– electrocyclic ring closing
conrotatory
HH
≡
H
H
H
6 e– electrocyclic ring closing
disrotatory
H
H H
≡
H
H
Chapter 4 15
H
4 e– electrocyclic ring closing
conrotatory
H
H H
≡
H
H
When small rings are fused to other rings, the cis ring fusion is almost always much more stable than the
trans ring fusion. The opposite is true only for saturated 6-6 or larger ring systems. (Make models to
confirm this.) The order of stability of the three possible products shown above is: cis-6-5 > trans-7-4 >
trans- 8-3.
4. (a) Chair TS, with the Me on the C(sp3) equatorial.
O
Ph
O
Ph
CH3
H3CH3C
CH3
≡O
PhH3C
CH3
(b) Chair TS, with the Ph equatorial.
O
Ph
O
Ph≡
OSiR3
OBn
OSiR3
O
PhOSiR3BnO BnO
(R)H
(S)
(c) Chair TS, with both substituents equatorial.
Ph
CH3
PhH3C
H
H
≡PhH3C
H
H
(d) Two different chairs are possible, but one (Ph equatorial) is lower in energy than the other.
Ph
CH3
PhH
CH3
H
≡Ph
H
CH3
HPh
H
H
CH3
(e) A chair TS is not possible, so it goes through a boat TS.
Chapter 4 16
HH3CO2C
CH3
O
H
H3C
HCH3H3CO2C
CH3
O
H
chair TS no way José
HH3CO2C
CH3
O
H
H3C
H
≡H3C
OH
H3CO2C
CH3
H
boat TS much better
(f) Again, a boat TS is necessary.
H
H
H60 °C
H
H
HH
≡
(g) A chair TS would produce a trans double bond in the seven-membered ring, so the boat TS is
operative, and the H and OSiR3 groups on the two stereogenic atoms are cis to one another.
OSiR3
HOSiR3
H
H
HOSiR3
HOSiR3
H
H
H
trans double bond
OSiR3
H
H
HOSiR3
H
H
H
OSiR3
OSiR3
H
≡
OSiR3H OSiR3H
(h) The chair TS is enforced in this macrocyclic compound.
Chapter 4 17
CH3
O
H3C
CH3
H3C
H
O
HCH3
O
O
HH
H3C≡
CH3CH3
O
O
HH
H3C
5. (a) First step: hetero-Diels–Alder reaction (six-electron, [4+2] cycloaddition). Second step: Claisen
rearrangement (six-electron, [3,3] sigmatropic rearrangement).
(b) The diene is electron-rich, so it requires an electron-poor dienophile for a normal electron demand
Diels–Alder reaction. The C=C bond of ketenes is pretty electron-rich, due to overlap with the lone pairs
on O: H2C=C=O ↔ H2C–
–C≡O+
. Only the C=O bond of the ketene is of sufficiently low energy to react
with the diene at a reasonable rate.
(c) First, it is important to remember that in ketenes, the p orbitals of the C=O bond are coplanar with the
substituents on the terminal C.
RS
RL
O
Because of the ketene’s geometry, in the TS of the hetero-Diels–Alder reaction, either RS or RL must point
directly at the diene. The lower energy approach towards RS is chosen, and the product in which RS
points back toward the former diene portion of the compound is obtained.
C
O
RSRL
O
RS
RL
+C
ORS
RL
Second step: The new σ bond forms between the bottom face of the double bond on the left and the
bottom face of the double bond on the right, giving the observed, less thermodynamically stable product.
Chapter 4 18
O
RS
RLH
HH O
RS
RLH H
HH O
RS
RLH H
HH
≡
O
RS
RL
H
H
H
H
6. (a) Number the C’s. C1, C2, C5 and C6 are clear in both starting material and product. The rest
follows.
H3C
O
OSiMe3
∆
H3C
Me3SiO
O
H
1 23 4
56 7
8 9 5
63
21
4
78
9
We break the C4–C6 bond, and we form C3–C8 and C4–C9. The formation of the latter two bonds and
the fact that we’re forming a cyclobutanone suggests a [2+2] cycloaddition between a ketene at C3=C4=O
and the C8=C9 π bond. We can generate the requisite C3=C4 π bond by electrocyclic ring opening of the
cyclobutene ring in the S.M.
H3C
O
OSiMe3
∆
Me3SiO
C OH3C
Me3SiO
H3CC
O
(b) Electrocyclic ring closing followed by base-catalyzed tautomerization (both starting material and
product are bases) gives the product.
NH
N
MeNSMe
N
O
NH
N
MeNSMe
N
H
O
B
NH
N
MeNSMe
N OHNH
N
MeNSMe
N O–
H B
Chapter 4 19
(c) Diels–Alder reaction followed by spontaneous elimination of Me3SiO– and aromatization gives the
product. Loss of Me3SiO– occurs so readily because the Me3Si group is a π electron withdrawer like a
carbonyl group.
OSiMe3
ArO
Me3SiO
CO2MeCO2Me
OSiMe3
ArO
Me3SiO
CO2MeArO
Me3SiO
H H–OSiMe3
CO2MeArO
Me3SiO
work-upCO2MeArO
HO
(d) The key atoms for numbering the C’s are C1 (with the 2-bromoallyl group attached), C7 (ester group
attached), and C8 (O attached). We form bond C1–C9 and break bond C3–C7. Since C3–C7 is the
central bond of a 1,5-diene system terminating in C1 and C9, i.e. C1=C2–C3–C7–C8=C9, this must be a
Cope rearrangement.
78
9
5
6
3
2
1
4 7
89
BrCO2Me
OSiMe3 ∆
Br
CO2Me
O
H
1
23
4
5 6
Br
CO2Me
OSiMe3 Br
CO2Me
OSiMe3 ≡
Br
CO2Me
OSiMe3
H
aq. workup
Br
CO2Me
OH
H
Br
CO2Me
O
H
(e) Numbering the carbons is made easier by C9, C8, and C4. These atoms make it easy to label C4
through C9. Since C11 is a carbanion, we can expect that it will add to C4, the only electrophilic C in the
starting material, and since C11 has a CH3 group attached, we can identify it and C10 in the product as the
Chapter 4 20
easternmost C’s, with C11 attached to C4. For C1 to C3, we preserve the most bonds if we retain the
C9–C3–C2–C1 sequence. So overall, we form C4–C11, C4–C2, and C10–C1, and we break C4–C3.
7
8
9
56 4
7
8
912
34
56
H3C
Me3SiO
O
H
Li ;
warm to RT;NaHCO3
CH3CH3
OH
OHH CH3
1011
11
3 21
10
The first step is addition of C11 to C4. We still need to form C10–C1 and break C4–C3. Since we have a
1,5-diene (C11=C10–C4–C3–C2=C1), we can do an oxy-Cope rearrangement. This gives a 5-8 system
in which we only have to form the C4–C2 bond. C4 is neither nucleophilic nor electrophilic, while C11 is
nucleophilic (conjugation from OSiMe3). Upon quenching with water, however, C4 becomes an
electrophilic carbonyl C, whereupon C11 attacks with concomitant desilylation of O to give the product.
H3C
Me3SiO
O
H
CH3 H3C
Me3SiO
H
CH3O–
1 10
4
11
43Li
H3C
Me3SiO
H
CH3O– ≡
CH3
Me3SiOH
O–H CH3
H OH
4
2
CH3
OH
OH CH3
Me3Si
HO–
CH3
OH
O–H CH3
CH3
OH
OHH CH3
(f) It’s clear that we form C4–C5 and C1–C6 bonds, and we break C1–C4. The strained C1–C4 bond
can be opened by an electrocyclic ring opening to give an o-xylylene, which undergoes an [8+2]
cycloaddition to give the observed product.
Chapter 4 21
Me3Si
Me3Si
CH3O
H
Me3Si
Me3Si
CH3O
HH
H
12
3 45
6 2
3 4
5
61
Me3Si
Me3Si
CH3O
H
Me3Si
Me3Si
CH3O
H
Me3Si
Me3Si
CH3O
H
H
H
(g) We form C2–C11 and C5–C9 bonds, and we eliminate the elements of Me3SiO2CCF3. The ZnCl2 is
a Lewis acid, so it coordinates to the carbonyl O and causes the cleavage of the carboxylate–C11 bond to
give the nice stable allylic cation C9–C10–C11. This cation can undergo a six-electron, [4+3] cycloaddi-
tion with the C2=C3–C4=C5 diene to give a new carbocation at C11. Loss of the Me3Si+ group from
C12 then gives the product.
Me3Si
O2CCF3
H3C CH3
ZnCl2
H3C
H3C
H1
2
3 4
5 67
89
1011
12
12
3 4
5
6
7
8910
11
12
Me3Si
O
H3C CH3CF3
OZnCl2
H3C
Me3Si
O
H3CO
F3C
Cl2Zn
Chapter 4 22
H3C
Me3Si
H3C
H3C
Me3Si
H3C
H3C
H3C
H
(h) The first product is formed by a hetero-ene reaction, with transfer of the H attached to S to the terminal
C of styrene.
O S
CH2
Ph
O S
Ph
CH3
H
H
H
The second product must incorporate two equivalents of the enol ether. We form C3–C5, C5–C4', and
C5'–S1 bonds, and we transfer a H from S1 to C4. A hetero-ene reaction forms the C3–C5 bond and
transfers the H. As for the other two bonds, since S1 and C5 are at the ends of a four-atom unit, we might
expect a Diels–Alder reaction. We can get to the requisite diene by eliminating the elements of BuOH by
an E1cb mechanism. The hetero-Diels–Alder reaction gives the product with endo stereoselectivity and the
expected regioselectivity.
O SH
OBu
O S OBu
CH3
12
3 45
12
34'
5'
4
5
O S
CH2
OBu
O S
CH3
OBu
H
H
H B
O S–
CH3
OBu
O S
CH3
OBu O S OBu
CH3
(i) We form C9–C1 and C4–C8 bonds, and we break C1–S and C4–S bonds. Since C1 and C4 are the
ends of a four-carbon unit, we can expect a Diels–Alder reaction. The cyclohexene in the product should
also tip you off. We can obtain the requisite diene by doing a [4+1] retro-cycloaddition, eliminating SO2
to give the C1=C2–C3=C4 diene. Stereospecific and endo-selective Diels–Alder reaction then gives the
Chapter 4 23
product.
SO2
NPh
O
BnN
O
Bn
H
HPh
1
2 3
45
6 78
99 8
7
65
4
3
2
1
SO2
NPh
O
Bn
NO
BnHPh
H NO
Bn
H
HPh
(j) When an acyl chloride is treated with Et3N, β-elimination takes place to give a ketene. When a
sulfonyl chloride is treated with Et3N, β-elimination takes place in the same way. The intermediate
undergoes [2+2] cycloadditions just like ketenes do to give the saturated four-membered ring.
H
H
H
S
O
O
ClEt3NH
H
S
O
O
analogous to ketene
SO2
N
H
H
H
S
O
O
N
H
(k) The second product provides the key. It is a six-membered ring with a single double bond, probably
the product of a hetero-Diels–Alder reaction. The requisite diene can be made from the starting material by
a vinylogous β-elimination, with NPhth as the leaving group. The same diene intermediate can undergo a
hetero-ene reaction to give the other observed product. An alternative mechanism for formation of the first
product, i.e. direct attack of the alkene (nucleophile) on S (electrophile, NPhth as leaving group) to give a
carbocation, followed by loss of H+, is also possible, but is less likely, especially since we know the C=S
compound is formed under the conditions. If this mechanism were operative it’s also likely that H+ would
be lost from the other C of the carbocation to give the more substituted and more stable isomeric alkene.
Chapter 4 24
H3C OCH3
SNPhth
O OH
pyrH3C OCH3
S
O O
H3C
CH3O2C
O
S
PhH3C
S
OH3C
CH3O2C
Ph
CH3
H3C
CH3O2C
O
S Ph
H O
CH3O2C
CH3
S Ph
H pyr
–O
CH3O2C
CH3
S Ph
HO
CH3O2C
CH3
S Ph
Hpyr
(l) Whenever you see a five-membered heterocycle, think 1,3-dipolar cycloaddition. The heterocyclic
rings shown can be made from an intramolecular cycloaddition of a nitrone and the alkene. The nitrone
must be made from the hydroxylamine and formaldehyde.
NH
OHH2C O N
OH
H O–
~H+
NO
OH
HN
O H
CH2–OH
H2C
O
N
O
N ANDO
CH2
NO
N
(m) Make: C5–O10, C5–O9, C6–O8, C6–O11. Break: C5–C6, C9–O10.
CH3
CH3
HO
H3C
O
CH3
CH3
OO
HO
H3C
1
2
3
4
5
6 16
54
3 2
7 7
O O O–
8
910
118
9
1011
Ozone lives to do 1,3-dipolar cycloadditions. After the cycloaddition to give the C6–O11 and C5–O9
Chapter 4 25
bonds, retro 1,3-dipolar cycloaddition occurs to break the C9–O10 and C5–C6 bonds. Then O8 can attack
C6 and O10 can attack C5 to give the observed intermediate (after proton transfer).
9 CH3
CH3
HO
H3C
O OO CH3
CH3
HO
H3C
OO
OH
H
CH3
CH3
OO
OH
H OH
H3C
56
6
10 11
56
9
10 11
8
O
CH3
CH3
OO
–O
H3C
H~H+ O
CH3
CH3
OO
HO
H3C
CH3
CH3
OOO
H
H OH
H3C
5
10
Second step. The elements of CH4O3 are eliminated. The most likely by-products are H2O and HCOOH.
Make: None. Break: C4–C5, C6–O8, O10–O11. The base can deprotonate the OH on C5, and the lone
pair on O can then push down to form a π bond with C5, causing the C4–C5 bond to break. The electrons
keep getting pushed around until they end up on O again and the O–O bond is broken, providing the
driving force for the step. A keto-aldehyde and formate anion are obtained. Now C7 (deprotonated) is
nucleophilic and C6 is electrophilic, so an aldol reaction followed by dehydration gives the observed
product.
O
CH3
CH3
OO
HO
H3C
aq. NaOH
CH3
CH3
O
16
54
3 2
1
2
3
4
7
7
611
10
9
8
8
O
OH5
10
9H2O
11
O
CH3
CH3
OO
O
H3C
H
HO–
O
CH3
CH3
OO–
O
H3C
H
HH
HH2O
7
6
CH3
CH3
O H
O
H
HO–
CH3
CH3
O H
–O
CH3
CH3
H
O
–OH
H ~H+
Chapter 4 26
CH3
CH3
H
–O
HO
HCH3
CH3
O
H
H
(n) Make: O9–C3. Break: C1–C3. Since O9 is nucleophilic, we must turn C3 into an electrophilic center.
ClCl OSiR3
OSiR3
Cl
O
OSiR3
AgNO3
acetone-H2O
H
1
2 3
4
5
6
78
9
101
23
4
5
6
78
910
In the first step, Ag+ promotes the departure of Cl– to give a cyclopropyl carbocation. This undergoes
two-electron disrotatory electrocyclic ring opening to give the chloroallylic cation, in which the empty
orbital is localized on C1 and C3. Then O9 can add to C3; desilylation then gives the product.
ClCl OSiR3
OSiR3
H
Ag+
– AgCl
Cl OSiR3
OSiR3
H HOSiR3
Cl OSiR3
Cl
O
OSiR3
H
SiR3
Cl
O
OSiR3
H
(o) The product is a 1,5-diene, specifically a γ,δ-unsaturated carbonyl, suggesting a Claisen rearrange-
ment. Work backwards one step from the product.
CH3
CH3 OH CH3(H3C)2N CH3
H3CO OCH3
CH3
CH3 CH3
12
34
5
978
(H3C)2N
O
78
91
23
45
6 6
Chapter 4 27
CH3
OMe2N CH3
CH3
6
8
91
2
3
CH3
OMe2N CH3
CH3
68
91
2
3
The immediate precursor retains the O6–C3 bond and would have a C8–O6 bond and a C8=C9 π bond.
This calls for an SN1 substitution at C8 to replace the C8–OMe bond with a C8–O6 bond and an E1
elimination to make the C8=C9 π bond. The overall reaction is an orthoamide Claisen rearrangement.
Me2N CH3
H3CO OCH3
Me2N CH3
OCH3
CH3
CH3 OH CH3
CH3
CH3 O CH3
Me2N
H3COCH3
H–OCH3
CH3
CH3 O CH3
Me2N
H3COCH3
CH3
CH3 O CH3
Me2N
H–OCH3
CH3
H3C O CH3
NMe2
H CH3
H3C O CH3
NMe2
H
(p) Make: C1–C9, C2–N6.
HN
O
OMe
OMeBr2 LDA
N
O
OMeMeOH
H 1
2
345
67
8
9
12
3
45
67
89
H
H
Since N6 and C9 are at the ends of a four-atom chain, we might expect a Diels–Alder reaction. The dieno-
phile in such a reaction would be benzyne; the key is the benzene ring fused to the new six-membered ring
and the fact that the H on C1 is gone in the product. (You could alternatively draw the π bond of the aro-
matic ring participating in the Diels–Alder reaction, but this is unlikely, because the π bonds of aromatic
rings are very bad dienophiles.) The first equivalent of LDA deprotonates N to make the 1,3-diene across
N6=C7–C8=C9; the second equivalent induces an E2 elimination across C1–C2 to give an aryne. Cyclo-
addition gives the enolate, which is protonated on C8 to give the observed product. In fact, this com-
pound is not very stable, and it is oxidized by air to give the fully aromatic product.
Chapter 4 28
N
O
OMe
OMeBr
H
H
H–Ni-Pr2
N–O
OMe
OMeBr
H
H
i-Pr2N–
N–O
OMeMeO
N–O
OMeMeO
H
H
work-up
N
O
OMeMeO
H
H
(q) Break: N3–N4, N4–C5. Make: C1–C6, N3–C5. We lose the elements of NH3.
NH
N
MeO
Ph
H
ZnCl2
NH
MeO Ph1
23
45
6
6
5
1
23 + NH3
4
H
HHH
Since we are forming a σ bond at the end of a six-atom chain and breaking the σ bond in the middle, we
might expect a Cope rearrangement. To do this, we must make a C5=C6 π bond. We can do this by
transposing the N4=C5 π bond. This transposition converts an imine to an enamine, which is exactly
analogous to converting a ketone to an enol. The enamine then undergoes Cope rearrangement to give the
C1–C6 bond. (Note how this Cope rearrangement is analogous to the Claisen rearrangement of O-allyl-
phenols.) After reestablishing aromaticity by tautomerization, nucleophilic N3 attacks electrophilic C5 to
form the N3–C5 bond. Finally, E1 elimination of NH3 gives the indole.
ZnCl2
NH
N
MeO
Ph
H
HHH
NH
N
MeO
Ph
H
HZnCl2 H
H
NH
N
MeOH
Ph
H
H
ZnCl2
H+
NHN
MeOH
Ph
H
H
ZnCl2
H+
Chapter 4 29
NH2
N
MeOH
Ph
H
ZnCl2 NH2
H
MeO
Ph
NZnCl2 H+
H H
NH2
H
MeO
PhHN
ZnCl2
H
NH2
MeO Ph
NH
H
H
ZnCl2
~H+
NH
MeO Ph
NH2
H
H
ZnCl2NH
MeO PhH
H NH
MeO Ph
H
(r) Make: C2–C4, C1–C3. Break: C1–C2. Since only one equivalent of malonate is incorporated into the
molecule, the other equivalent must act as a base. The migration of C1 from C2 to C3 is a 1,2-alkyl shift.
Under these basic conditions, it is likely to proceed by a Favorskii mechanism. Deprotonation of C3 by
malonate gives the enolate. Two-electron electrocyclic ring closing with expulsion of Cl– gives the
cyclopropanone. Attack of malonate on C2 gives a tetrahedral intermediate; fragmentation of this with
expulsion of Cl– gives the observed product. Other reasonable mechanisms can be drawn, some of which
do not involve an electrocyclic ring closing.
H3C Cl
H3C Cl
O
EtO2C CO2Et
Na
2H3C
CO2Et
CH3 O
CO2Et
12
3 4 42
31
H3C Cl
H3C Cl
O
H H
CO2Et
CO2Et
Cl Cl
H3C CH3
O–
H
O
H3C
H3C
Cl
H
CO2Et
CO2EtH3C
H3C
Cl
H
–O CO2Et
CO2EtH3C
CO2Et
CH3 O
CO2Et
(s) The five-membered heterocycle should alert you to a 1,3-dipolar cycloaddition.
Chapter 4 30
H3C
O H+
H3C
OH
N CH3
HO
HH3C
OHNH3C
OHH
~H+
H3C
OH2NH3C
OH
H3C
NH3C O
H ON
H3C
H3CO
N
H3C
H3C
(t) Make: C1–C9, C2–C6, O6'–C9. Break: C9–N2.
N
O O
EtO
CO2MeN2MeN
NMe
N
O
CO2Me
Et
OH
O
cat. Rh2(OAc)41
23
45
67
8
910
12
4' 6'
8'
34
4' 5
6
6'
7
8 8'9
10
C6 and C9 are at opposite ends of a four-carbon unit, but since one of these atoms (C7) is saturated and
quaternary, a Diels–Alder reaction is unlikely (can’t make diene). The combination of a diazo compound
with Rh(II) generates a carbenoid at C9. The nucleophile O6' can add to the empty orbital at C9, gener-
ating the O6'–C9 bond and a carbonyl ylide at C6–O6'–C9. Carbonyl ylides are 1,3-dipoles (negative
charge on C9, formal positive charge on O6', electron deficiency at C6), so a 1,3-dipolar cycloaddition
can now occur to join C2 to C6 and C1 to C9, giving the product. Note how a relatively simple tricyclic
starting material is transformed into a complex hexacyclic product in just one step!
cat. Rh2(OAc)4N
O O
EtO
CO2MeMeN
9
N
O
EtO
CO2Me6'9
6
N2
NMe
N
O
CO2Me
Et
OH
O
9
6'
62
1 NMe
N
O
CO2Me
Et
OH
O
12
6
6'
9
Chapter 4 31
(u) The cyclobutanone should tip you off to a ketene–alkene cycloaddition. Ketenes are generally made
by Et3N-catalyzed elimination of HCl from acyl chlorides. Oxalyl chloride ClCOCOCl serves to convert
the acid into an acid chloride.
OR
O
OH
Cl
O
OR
O
OH
–O Cl
COCl
OR
O
OH
COCl
O
Cl–COCl
OR
O
OH
O
Cl
OCl
OR
Cl
OH
CO2 CO
Cl– O
Cl
O
CH3
O
C
CH3
O
O
HCH3
Cl
O
H
H NEt3CH3
O
O
H
(v) Another five-membered heterocycle, another 1,3-dipolar cycloaddition. The first step is formation of
the requisite 1,3-dipole, a nitrile ylide, by a two-electron electrocyclic ring opening. Then dipolar
cycloaddition occurs.
H3C
NPh
N
CH3
Ph
NCH3
Phhν
(w) Formally this reaction is a [2+2] cycloaddition. In practice, concerted [2+2] cycloadditions occur
under thermal conditions only when one of the components is a ketene or has a π bond to a heavy element
like P or a metal. Neither of the alkenes in this reaction fits the bill. However, one of these alkenes is
very electron rich and the other is very electron poor, so a nonconcerted, two-step polar mechanism is
likely.
H3C
CH3
NMe2
PhNO2
H3C
CH3
Ph NO2
NMe2
∆H3C
CH3
Ph NO2
NMe2
Chapter 4 32
(x) The extra six C’s must come from benzene. A photochemically allowed [2+2] cycloaddition between
the alkyne and benzene gives an intermediate that can undergo disrotatory electrocyclic ring opening to
give the observed product (after bond alternation). (Either two or three arrows can be drawn for the
electrocyclic ring opening, but the TS for the reaction involves all eight π electrons, so to be disrotatory the
reaction must be promoted photochemically.) Benzene does not usually undergo cycloaddition reactions,
but here it evidently does.
hνCO2CH3
CO2CH3CO2CH3
CO2CH3 CO2CH3
CO2CH3
H
H
H
H
H
H
(y) The second product is clearly obtained by a hetero-Diels–Alder reaction between acrolein and
isobutylene. The first product is less obvious. Two new C–C bonds are formed, and H atoms are
transferred from C7 and C8 to C2 and O4. This suggests two ene reactions.
O +CH3
CH3
HO
+
OCH3
CH3
300 °C12
34 5
6
7
812
3
4
5
6
7
8
4
3
2 1 5 or 8
67
8 or 5
OH3C CH3 O
CH3
CH3
O
H3CH
O
H3CH H ≡ H
O
H
H
O
H ≡HO
(z) Elimination of allyl alcohol occurs by an E1 mechanism. Then a Claisen rearrangement gives the
product.
Chapter 4 33
O
O
H+
O
O
H
O
HH
O O
(aa) The C6 to C9 unit in the product is numbered by virtue of the two H’s on C6. Make: C2–C9, C3–
C6, C7–O11. Break: C9–S10, S10–O11.
MeN S
O–
Ph A+
MeNHO
H
Na2S
CH3OH
∆1 2
34
56
78
9 10
11 1
2
3
45
6
7
8 9
11
Formation of C2–C9 and C3–C6 suggests a Diels–Alder reaction, this one of the inverse electron demand
flavor. The regioselectivity follows the ortho-para rule and the stereoselectivity is endo. The C7–O11
bond can now be formed and the C9–S10 bond cleaved by a [2,3] sigmatropic rearrangement to give
compound A. All that is left is to cleave the S10–O11 bond. Na2S attacks S, with RO– acting as the
leaving group, and protonation gives the final product.
MeN
S
H–O Ph
NMe
H
S–O
PhMeNO
H
PhS
A
NaS–
MeN–O
H
MeO HMeNHO
H
(bb) Retro Diels–Alder reaction gives off N2 and an ortho-xylylene. With no other substrates available,
this extremely reactive substance dimerizes in another Diels–Alder reaction to give the product.
Chapter 4 34
N
N
(cc) The product is formally the result of a [1,3] sigmatropic rearrangement. STOP! [1,3] sigmatroopic
rearrangements are very rare, and they should be viewed with suspicion. They are thermally allowed only
when one of the components is antarafacial. Sometimes an apparent [1,3] shift is actually the result of two
sequential reactions (polar or pericyclic). In this case, the presence of KH suggests an oxyanion-acceler-
ated concerted process. The one-atom component can be antarafacial if the back lobe of the sp3 orbital
used to make the old bond to C6 is then used to make the new bond to C4. After workup, aromaticity is
reestablished by protonation-deprotonation.
O
OH
KH
O
OH1
2
34
5
67
8
9
1
2
34
56
7
89
O H
HH
O–
O HH
H
–O
O HH
H
–O
O
OHH
H+ O
OHH
O
OH
H+ work-up
(dd) Make: C1–C11, C5–S13, C6–C10, O12–S13. Break: C7–O12, S13–Cl.
CH3
OH
H3CCH3
PhSCl
Et3N
S
CH3
CH3
CH3
–O Ph
1
2
34 5 6 7 8
9
10 11
12
3 45 6 7
8
9101112
13
1213
The product looks very much like the result of a Diels–Alder reaction that forms the C1–C11 and C6–C10
bonds. Work backwards one step from the product.
Chapter 4 35
S
CH3
CH3
CH3
–O PhS
CH3
CH3
CH3
–O Ph
The intermediate might be made by a [2,3] sigmatropic rearrangement of an RO–SPh compound.
CH3
O
H3CCH3
H
NEt3
CH3
O–
H3CCH3
PhS Cl
CH3
H3CCH3
OPhS
S
CH3
CH3
CH3
–O PhS
CH3
CH3
CH3
–O Ph
(ee) Make: C6–C7, C3–C8. Break: C5–C6, C3–O4.
O
O
ONaH CO2EtEtO2C
CO2Et
CO2Et
OH
1
23 4
56 7 8 1
23
67
8
The two new bonds can be obtained by a Diels–Alder reaction. First, deprotonation gives an enolate that
has an ortho-xylylene resonance structure. Diels–Alder reaction followed by retro-Diels–Alder reaction
gives the product.
CO2Et
EtO2C
O
O
O–H
H H
OO
O–
H
O–O
O
H
Chapter 4 36
CO2EtEtO2C
O–O
O
H
–O
CO2Et
H
CO2Et
work-up
CO2Et
CO2Et
OH
(ff) As in the previous problem, Diels–Alder reaction followed by retro-Diels–Alder reaction establishes
the desired C–C bonds. Then E1 elimination of CH3OH gives the desired product. (An E1cb mechanism
for elimination is also reasonable, but less likely in the absence of strong base.)
O
CO2CH3
O
OCH3
OCH3
O
CO2CH3
O
OCH3H3CO
CO2CH3
OCH3
OCH3
H
H
CO2CH3
H
H OCH3
–OCH3
CO2CH3
OCH3
H
(gg) The D atoms give major clues to the numbering. Break: C5–C6, C9–C10, C11–C12. Make: C5–
C11, C10–C12.
hνD
D
D
D
1 2
3
45 6
7
8910
2
34
56
7
8910
11
121112
1
If we break the C5–C6 and C9–C10 bonds by a retro-Diels–Alder reaction first, we get two molecules of
benzene. But irradiation of benzene doesn’t give the observed product, so this can’t be right. Instead,
let’s form the C5–C11 and C10–C12 bonds first by a (photochemically allowed) [2+2] cycloaddition.
This gives the strained polycyclic compound shown. Now the C5–C6 and C10–C9 bonds can be broken
by a [4+2] retro-cycloaddition (thermal, supra with respect to both components) to give the tricyclic
compound. This compound can then undergo disrotatory six-electron electrocyclic ring opening (thermal)
to give the observed product. Note that only the first reaction in this series requires light.
Chapter 4 37
hν
D
D
D
D
5
1011
12D
D
≡H
H
H
H
D
D
H
H
D
D
(hh) Numbering the product is difficult. Because C9 in the starting material has no H atoms, let’s make it
one of the C’s in the product that has no H atoms. Make: C1–C9, C2–C9, C5–C9. Break: C1–C2.
±C±1
2
345
6
7 89
H
H
H
H
H
H
H H
3
4
91
2
56
7
8
Carbenes like to do [2 + 1] cycloadditions to alkenes. Such a cycloaddition between C9 and the C1=C2 πbond gives a product which can undergo an 8-electron electrocyclic ring opening to cleave the C1–C2
bond, then a 6-electron electrocyclic ring closing to form the C5–C9 bond. All that is left to do is a [1,5]
sigmatropic rearrangement to move the C5 H to C4.
±C±
H
H
H
H
H
H
H
H
±
H
H
H
H
±H
H
H
H
H
H
H
H
≡
H
H
H
H
H
H
H
H
H
H
H
H
(ii) Following the instructions, we number all the N atoms. The byproducts are 2 N2. Make: C2–O4,
Chapter 4 38
C3–N9, N6–N9. Break: C2–N9, N6–N7, N9–N10.
EtO2C
N N
Me
O
∆
N N
OEtO2C Me
NN N N
12 3
4
5
678 9 10 11
12 3
45
6 9
The thermal reaction that azides undergo is the Wolff rearrangement (Chapter 2). In the present case, the
Wolff rearrangement allows us to make the C3–N9 bond and cleave the N9–N10 bond. A resonance
structure can be drawn in which N9 has a negative charge. This lone pair is used to attack N6, displacing
N2. Next, the C2–N9 bond is cleaved by a 4-electron electrocyclic ring opening to give a nitrilimine,
which then undergoes a 6-electron electrocyclic ring closure to give the product.
EtO2C
N N
Me
O
NN N N
∆ EtO2C
N N
Me
O
NN
EtO2C
N N Me
O
NN
EtO2C
N N Me
O EtO2C
N N Me
O EtO2C
N NMe
O
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